I'm working through a linear algebra book and in the appendix of the book they state the following,
"The principal square root of $A$ is defined to be $$A^{\frac{1}{2}}=V D^{\frac{1}{2}}V^T$$ where $A$ is a positive semi-definite and symmetric matrix, $D$ is a diagonal matrix and $V$ is an orthogonal matrix. The matrices hold the following, $$A^{\frac{1}{2}} (A^{\frac{1}{2}})^T=A$$ and $$(A^{\frac{1}{2}})^T A^{\frac{1}{2}}=A$$ provided that $A$ is invertible."
I've tried proving this using the definition of a square root matrix via diagonalization by using $$ \begin{align} A^{\frac{1}{2}} A^{\frac{1}{2}}&= (V D^{\frac{1}{2}}V^T)(V D^{\frac{1}{2}}V^T) \\ &=V D^{\frac{1}{2}} (V^TV) D^{\frac{1}{2}}V^T \\ &=V D^{\frac{1}{2}}D^{\frac{1}{2}}V^T \\ &=VDV^T \\ &=A \end{align} $$ I do note that since $A$ is symmetric then $V^T=V^{-1}$ as the proofs above are just modifications of proving the square root of a matrix. However, I'm having some issue understanding how to implement the transpose properly for either case,
$$ \begin{align} (A^{\frac{1}{2}})^T A^{\frac{1}{2}}&= (V D^{\frac{1}{2}}V^T)^T(V D^{\frac{1}{2}}V^T) \end{align} $$
So my question is two parts,
- How does taking the transpose of $A^{\frac{1}{2}}$ affect the proof properly? As I believe there is something simple I am missing.
- Can the same proof be done to show that $A^{-\frac{1}{2}} (A^{-\frac{1}{2}})^T=A^{-1}$ and $(A^{-\frac{1}{2}})^T A^{-\frac{1}{2}}=A^{-1}$?
Thank you